JEE PYQ: Vector Algebra - Question ID 43173f04f8d9 (JEE Main 2024)

Core Formula & Concept:

When two vectors $\vec{A}$ and $\vec{B}$ of equal magnitude $R$ are inclined at an angle $\theta$, their sum and difference can be found using the following fundamental results from vector algebra:

• Magnitude of the Sum: $|\vec{A} + \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 + 2|\vec{A}||\vec{B}|\cos\theta}$ Since $|\vec{A}| = |\vec{B}| = R$, this simplifies to: $|\vec{A} + \vec{B}| = \sqrt{R^2 + R^2 + 2R^2\cos\theta} = \sqrt{2R^2(1 + \cos\theta)}$
• Magnitude of the Difference: $|\vec{A} - \vec{B}| = \sqrt{|\vec{A}|^2 + |\vec{B}|^2 - 2|\vec{A}||\vec{B}|\cos\theta}$ Again, using $|\vec{A}| = |\vec{B}| = R$: $|\vec{A} - \vec{B}| = \sqrt{2R^2(1 - \cos\theta)}$
• Trigonometric Identities: To simplify the expressions involving $1 \pm \cos\theta$, we use the half-angle identities: $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right), \quad 1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$

Step-by-Step Derivation:

Step 1: Compute $|\vec{A} + \vec{B}|$

$|\vec{A} + \vec{B}| = \sqrt{R^2 + R^2 + 2R^2\cos\theta} = \sqrt{2R^2(1 + \cos\theta)}$ Using the identity $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right)$: $|\vec{A} + \vec{B}| = \sqrt{2R^2 \cdot 2\cos^2\left(\frac{\theta}{2}\right)} = \sqrt{4R^2\cos^2\left(\frac{\theta}{2}\right)} = 2R\left|\cos\left(\frac{\theta}{2}\right)\right|$ Since $\theta$ is the angle between two vectors, it lies in the range $0 \leq \theta \leq \pi$, so $\frac{\theta}{2}$ lies in $\left[0, \frac{\pi}{2}\right]$, and $\cos\left(\frac{\theta}{2}\right) \geq 0$. Thus: $|\vec{A} + \vec{B}| = 2R\cos\left(\frac{\theta}{2}\right)$ This matches Option A.

Step 2: Compute $|\vec{A} - \vec{B}|$ (for completeness)

$|\vec{A} - \vec{B}| = \sqrt{2R^2(1 - \cos\theta)}$ Using the identity $1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$: $|\vec{A} - \vec{B}| = \sqrt{2R^2 \cdot 2\sin^2\left(\frac{\theta}{2}\right)} = \sqrt{4R^2\sin^2\left(\frac{\theta}{2}\right)} = 2R\left|\sin\left(\frac{\theta}{2}\right)\right|$ Again, since $\frac{\theta}{2} \in \left[0, \frac{\pi}{2}\right]$, $\sin\left(\frac{\theta}{2}\right) \geq 0$, so: $|\vec{A} - \vec{B}| = 2R\sin\left(\frac{\theta}{2}\right)$ This does not match any of the given options directly, but it confirms that Option C is incorrect (it has $\sqrt{2}R$ instead of $2R$).

Step 3: Verify Other Options

• Option B: Claims $|\vec{A} - \vec{B}| = 2R\cos\left(\frac{\theta}{2}\right)$. This is incorrect, as shown above.
• Option D: Claims $|\vec{A} + \vec{B}| = 2R\sin\left(\frac{\theta}{2}\right)$. This is incorrect, as the correct expression involves $\cos\left(\frac{\theta}{2}\right)$.

Common Traps & Exam Tip:

Students often make the following mistakes in this question:

1. Confusing Sum and Difference: Many students mix up the formulas for $|\vec{A} + \vec{B}|$ and $|\vec{A} - \vec{B}|$, leading to incorrect options like B or D. Remember:
   • $|\vec{A} + \vec{B}|$ involves $+2|\vec{A}||\vec{B}|\cos\theta$ and simplifies to $2R\cos\left(\frac{\theta}{2}\right)$.
   • $|\vec{A} - \vec{B}|$ involves $-2|\vec{A}||\vec{B}|\cos\theta$ and simplifies to $2R\sin\left(\frac{\theta}{2}\right)$.
2. Incorrect Trigonometric Identities: Some students forget the half-angle identities or misapply them, leading to expressions like $\sqrt{2}R\sin\left(\frac{\theta}{2}\right)$ (Option C). Always recall: $1 + \cos\theta = 2\cos^2\left(\frac{\theta}{2}\right), \quad 1 - \cos\theta = 2\sin^2\left(\frac{\theta}{2}\right)$
3. Sign Errors: While the absolute value ensures non-negative results, students sometimes overlook the range of $\theta$ and incorrectly assume $\cos\left(\frac{\theta}{2}\right)$ or $\sin\left(\frac{\theta}{2}\right)$ could be negative. For $\theta \in [0, \pi]$, both are non-negative.

Exam Tip: Always sketch the vectors $\vec{A}$ and $\vec{B}$ as two sides of a parallelogram. The sum $\vec{A} + \vec{B}$ is the diagonal that "adds" the vectors, while the difference $\vec{A} - \vec{B}$ is the other diagonal. This geometric intuition helps verify the algebraic results.